LeetCode OJ算法题（三十七）：Count and Say

题目;

The count-and-say sequence is the sequence of integers beginning as follows:
1, 11, 21, 1211, 111221, ...

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n, generate the nth sequence.

Note: The sequence of integers will be represented as a string.

解法：

连题目都已经按递推公式给出了，不知道怎么能不用递归实现呢？

首先当n=1时，直接返回“1”；

假设发f（k）已经求出，记为s，那么我们只需统计s中的重复字符即可，每一段重复的字符重新编码成”count“+“char”的形式，再依次拼接起来即可。

public class No37_CountAndSay {public static void main(String[] args){System.out.println(countAndSay(5));}public static String countAndSay(int n) {if(n == 1) return "1";String ret = "";        String s = countAndSay(n-1);        int index = 0;        int count = 0;        char pre = '\0';        while(index < s.length()){        if(s.charAt(index) != pre){        if(count > 0) ret = ret+count+pre;        pre = s.charAt(index);        index++;        count = 1;        continue;        }        count++;        index++;        }        ret = ret+count+pre;        return ret;    }}